This is a fun and whimsical mathematical application: an extension of the Manhattan grid systems to every location on Earth!
Find Brasilia at South 39,688 St and East 25,424 Ave, Bangalore at 84,097 St and East 77,224 Ave, or Bora Bora at the corner of 41,214 Ave and S 69,342 St.
I wonder how much the cab fare would be from midtown Manhattan to South 43,892 St and East 71,247 Ave?
While others celebrate the number 11 on this special day, I prefer to honor the Equilateral Triangle.
Last year, on 10/10/10, I celebrated the symmetry of the equilateral triangle. This year, I offer a favorite Proof Without Words. Well, a proof with some words. In any event, we will use equilateral triangles to prove that the following infinite series
is 
Consider the following diagram.
Notice that the largest blue equilateral triangle is
So, the sum of the blue triangles is
Let’s call this S.
Now, here’s the magic: the sum of the red triangles is also S! This is true because for every blue triangle, there is a congruent red triangle right next to it. Similarly, the sum of the yellow triangle is also S.
When you put all the blue, red, and yellow triangles together, you get the original triangle, whose area is 1. Thus, 3S = 1, and so
Therefore, we have
Happy Equilateral Triangle Day!
Related Posts
Today’s date, 11-10-11, reminded me to re-visit a recent post that posed the question “Which triangle is more equilateral: the 10-10-11 triangle, or the 10-11-11 triangle?”
The original post elicited lots of great comments from readers, who weighed in on what they thought the question meant and how they might go about trying to answer it. I offered one approach, and an answer to the question, in this follow up post.
As a math teacher, there are many reasons I try to create problems like this. Here are a few that I think are important.
First and foremost, in order to address this question, a significant amount of thought must be put into deciding what the question means. This process involves analysis, synthesis, reflection, and ideally discussion, all of which will be substantially mathematical in nature.
A second, related, virtue is that there is no obviously correct interpretation of what this question means. Mathematics is often viewed in stark terms: answers are either right or wrong. But the certitude of mathematics comes only after we agree on mathematical models for our given problem. There is often great debate about what those models should look like; the history of mathematics is full of such debate.
Problems like this one invite students into the modelling process, where they can discuss and debate the validity of various approaches. Moreover, the problem allows solvers to create multiple different models to explore, compare, and contrast. And in the end, we can pose and explore meta-mathematical question like “Which model most closely aligns with our intuitions?” and “Which model is the most useful?”
Lastly, this problem demonstrates one role creativity plays in mathematics. A simple response to the question, one I heard many times, is “Neither of these triangles are equilateral. They are both equally unequilateral.” Given our rigid definition of what equilateralmeans, this response is technically correct. But by relaxing our ideas about equilateral, by allowing ourselves to ponder what the phrase “more equilateral” might mean, by thinking creatively about what kinds of questions we can ask, we create an opportunity to explore, and possibly uncover, some new mathematical ideas.
Richard Geller, a longtime math teacher and math team coach at Stuyvesant High School, recently passed away. I only knew Richard professionally, but it was easy to see that he was a good man and a good teacher. His dedication to his students, his school, and the math team was always apparent.
At a math circle one evening, Richard shared with me a lovely solution to a challenging problem that I’ll never forget. I share it here as a tribute to him.
There are lots of famous concurrencies in triangles. The medians of a triangle all intersect at the centroid; the angle bisectors at the incenter; and the perpendicular bisectors at the circumcenter. We say that each set of lines is concurrent.
A less intuitive concurrency is that of a triangle’s altitudes, which all intersect at the triangle’s orthocenter. It’s harder to see because you often have to rethink your notion of altitude to see them intersect.
Not only is it harder to see the concurrency of the altitudes, but it’s harder to prove it as well. There are many well-known methods, like using Ceva’s Theorem or areas, but they are rather complicated. To me, the orthocenter was never as accessible as the circumcenter, incenter, or centroid. Until Richard showed me this proof.
Start with an ordinary triangle. We want to show that the altitudes of this triangle all intersect at a single point.
First, we create a new triangle by rotating three copies of our original around the midpoints of each side. What we are doing is creating a new triangle whose medial triangle is our original triangle.
Now the magic: construct the perpendicular bisectors of the new triangle.
The amazing fact here is that the perpendicular bisectors of the new triangle are the altitudes of the original triangle! As long as we know that the perpendicular bisectors of any triangle are concurrent (which is fairly easy to prove), we know that the altitudes of any triangle are concurrent, too!
Richard didn’t invent this theorem or this proof, but he taught it to me, and for that I’ll be forever grateful. When I share it with students, I think of him. And from now on, when I show students the Geller Technique, I’ll wrap it up with one of Richard’s favorite phrases: Math is #1!
After producing the third-lowest home run total in the National League, the New York Mets are altering the dimensions of their home park, Citi Field.
https://www.espn.com/new-york/mlb/story/_/id/7174665/new-york-mets-officially-announce-citi-field-dimension-changes
Only 108 home runs were hit at Citi in 2011. The reconfigured outfield, from left to right, will have new distances from home plate of 335, 358, 385, 408, 398, 375, and 330 feet. The old distances were 335, 371, 384, 408, 415, 378, and 330.
The Mets claim that this reduces the in-play surface by 2 percent. The area of play more than 300 feet from home plate will be reduced by 5 percent.
Comparing the areas of these regions is a fun little math problem. The real question, of course, is how many more home runs will be hit at Citi Field next year?
